Given Q=Q0e−Rt/2L R=1.5Ω,L=12mH,In(2)=0.693 Q=0.5Q0,t= ? 0.5Q0=Q0e−Rt/2L ⇒21=e−Rt/2L Taking log on both sides ln(21)=lne−Rt/2L ⇒ln2=2LRt t=R2LIn2=1.52×12×10−3×0.693 t=11.09 ms
The amplitude of the charge oscillating in a circuit decreases exponentially as Q=Q0e−Rt2L, where Q0 is the charge at t=0 s. The time at which charge amplitude decreases to 0.50Q0 is nearly: [Given that R=1.5Ω,L=12mH,ln(2)=0.693 ]
Held on 30 Apr 2024 · Verified 9 Jul 2026.
19.01 ms
11.09 ms
19.01 s
11.09 s
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